I am confused by the following problem in Aluffi's Algebra Chapter Zero...
A discrete valuation on a field $k$ is a surjective group homomorphism $v : k^* \to (\Bbb Z,+)$ such that $v(a + b) \geq \min(v(a),v(b))$ for all $a,b \in k^*$ with $a + b \in k^*$. Then the subring $R = \{ a \in k : v(a) \geq 0 \} \cup \{0\}$ is a Euclidean domain.
Here is what I am thinking:
Let $a, b \in k$.
If $a \not\mid b$ in $R$, then $v(a) \not\mid v(b)$. For $v(b) = q v(a) = v(a^q)$ implies $v(b a^{-q} ) = 1 \geq 0$, so $b a^{-q} \in R$, which means that $b a^{-q} a^{-q + 1} a = b$, so that $a\mid b$ in $R$.
Since $Z$ is a Euclidean domain and since $v$ is surjective, $v(a) \not\mid v(b)$ implies that there are integers $q$ and $v(r)$ such that $v(b) = q v(a) + v(r) = v(a^q r)$. However, this implies that $b r^{-1} a^{-q}$ is in $R$ which again implies that $a | b$. I am led to believe that $a | b$ for all $a$ and $b$, which is certainly not true...
Can someone help me see my mistake?
Thank you.
If $v(a)\ge v(b)$, then $ab^{-1}\in R$, so $a=bq$, $q\in R$. If $v(a)<v(b)$, then $a=b\cdot 0+a$.